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Transcript

ASTR 100: Homework 1 Solutions McGaugh, Fall 2008 1. Problem 1-15. “Our solar system is bigger than some galaxies.” This statement makes no sense. A galaxy is composed of hundreds of billions of stars, while our solar system is ‘just’ the stuff orbiting one star, our sun. Our solar system is a tiny part of the Milky Way Galaxy, being much smaller than it. 2. Problem 1-41. Distance by light: The problem asks you to compute the distance that light can travel in different intervals of time, measured in kilometers and miles. d = vt: distance = speed x time. In this case, the speed is the speed of light: v = c. According to Appendix A of the book, c = 3 × 105 km/s. The exact number is c = 299, 792 km/s; it is OK to use either value. a) 1 light-second = (3 × 105 km/s)(1 s). The seconds cancel out (see Appendix C), so 1 light-second = 300,000 km. 1 km = 0.62 miles, so (300,000 km)(0.62 miles/km) = 186,000 miles. That’s a long way to go in one second! b) Similarly, 1 light-minute = (3 × 105 km/s)(1 minute). Be consistent in the use of units. 1 minute = 60 seconds, so 1 light-minute = (3 × 105 km/s)(60 s). Now the units again cancel nicely, and 1 light-minute = 1.8 × 107 km = 1.1 × 107 miles. c) 1 light-hour = (3 × 105 km/s)(1 hour). 1 light-hour = (3 × 105 km/s)(60 minutes/hour)(60 seconds/minute) 1 light-hour = 1.08 × 109 km = 6.7 × 108 miles. d) 1 light-day. There are 24 hours in a day, so 1 day = (24 hours)(60 minutes/hour)(60 seconds/minute) = 86,400 seconds. 1 light-day = (3 × 105 km/s)(86, 400 s), so 1 light-day = 2.6 × 1010 km = 1.6 × 1010 miles. For a little perspective, the Voyager 1 spacecraft was launched 31 years ago, in 1977. It has been traveling away from the earth ever since, and is now the most remote man-made object at a distance of 1.6 × 1010 km. Light the sun emits today will pass Voyager tomorrow. 3. Problem 1-42. Moonlight and Sunlight: This simply inverts the problem above: Since d = vt, t = d/v. a) Light travel time from the moon to earth: 5 t = dmoon c . dmoon = 3.8 × 10 km (Textbook Table E.3). 5 km t = 3.8×10 3×105 km/s t = 1.3 seconds. b) Light travel time from the sun to earth: 8 t = dsun c . dsun = 1.5 × 10 km (Table E.2). 5 /s t = 3×10 km 1.5×108 km t = 500 s = 8 minutes, 20 seconds. 1 4. Problem 2-24. “If the Earth’s orbit were a perfect circle, we would not have seasons.” This statement is clearly false. A perfectly circular orbit would mean that the earth is always the same distance from the sun. However, the primary cause of our seasons is the tilt of the earth’s spin axis with respect to its orbital plane. Eliminating the variation in distance from the sun would not affect this primary attribute. 5. Problem 2-28. The north celestial pole (i.e., the North Star) is at an altitude of 35◦ above your northern horizon. This tells you that (a) you are at latitude 35◦ N. 6. Problem 2-32. If the sun rises precisely due east, (b) it must be the day of the equinox (either spring or fall). 7. Problem 3-13. “The Yankees are the best baseball team of all time.” On its face, this is the assertion of a partisan fan, not a scientific claim. However, can we evaluate this claim scientifically? Is there some measure by which the Yankees (or some other team) could be evaluated? Baseball is a statistics rich sport; there are many well-recorded possibilities. Which team has the best overall won-lost record? Which team has won the most pennants? Which has won the most world series? These are all questions that can be answered clearly and quantitatively. The trick is agreeing which statistic qualifies a team as the ‘best of all time.’ It often happens that even when we agree on the facts, we may not agree on what they mean. Does a team qualify as the best-ever if it meets one of the criteria above? Only if it meets all three? Or is there some other measure that would be more appropriate? The problem returns us to the phrasing of the assertion: what do we mean by ‘best?’ If we can agree on that, and can define a quantitative way to measure it, then we can turn this assertion into something accessible to science. Agreeing on what we mean by ‘best’ is not simple, which is why scientists prefer more specific, limited language. “The Yankees have been the most successful franchise in winning pennants” would be a more measured (and measurable!) statement. 8. Extra credit 2-44. The Sun’s diameter. θ = L/D. In this case, we want to find the size of the sun (L) from its observed angular diameter (θ = 0.5◦ ) and distance (D = dsun = 1.5 × 108 km). Hence L = θD. This formula holds only when θ is in radians, so first convert units: 1 radian = 180/π degrees. θ = 0.5◦ = 0.5 degrees(π/180)(radians/degree) θ = 0.0087 radians. Now L = (8.7 × 10−3 )(1.5 × 108 km) L = 1.31 × 106 km, very close to the actual value of 1.39 × 106 km. Indeed, we only miss it by a little because we have approximated the angular size of the sun as half a degree. This is close, but not exact. We can invert the procedure to find the exact value: θ = L/D = (1.39 × 106 km)/(1.5 × 108 km) = 0.0093 radians = 0.53◦ . 2